https/fedugen.w gen/lti/main.uni Halliday, Fundamentals of Physics, 10e Helo I S

Question

https/fedugen.w gen/lti/main.uni Halliday, Fundamentals of Physics, 10e Helo I S PRINTER VERSTON BACK Chapter 04, Problem 056 An Earth satellite moves in a circular orbit 717 km above Earth's surface with a period of 98.93 min. What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite? (a) Number Units (b) Number Units LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM VIDEO MINI-LECTURE MATH HELP MATH HELP MATH HELP MATH HELP LINK TO TEXT Question Attempts: 0 of 6 used SAVE FOR LATER SUBKET ANS nt/testiago stum?idrosnrnt2 132541 #N10024

Explanation / Answer

(a)

Since the radius of Earth is 6.37 *10^6m, the radius of the satellite orbit is

r= (6.37*10^6+ 717*^10^3) m = 7.08*^10^6m

Therefore, the speed of the satellite is

= 2**r/T = 2*(7.08*10^6)/(98.93min)*(60s/min) = 7.49*10^3m/s

(b) The magnitude of the acceleration is

av = 2/r = (7.49*10^3)^2/(7.08*10^6) = 7.923m/s^2

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